005 Sample Final A, Question 13

Question Give the exact value of the following if its defined, otherwise, write undefined.
$(a)\sin ^{-1}(2)\qquad \qquad (b)\sin \left({\frac {-32\pi }{3}}\right)\qquad \qquad (c)\sec \left({\frac {-17\pi }{6}}\right)$

Foundations:
1) What is the domain of $\sin ^{-1}?$
2) What are the reference angles for ${\frac {-32\pi }{3}}$ and ${\frac {-17\pi }{6}}$?
Answers:
1) The domain is $[-1,1].$
2) The reference angle for ${\frac {-32\pi }{3}}$ is ${\frac {4\pi }{3}}$, and the reference angle for ${\frac {-17\pi }{6}}$ is ${\frac {7\pi }{6}}$

Step 1:
For (a), we want an angle $\theta$ such that $\sin(\theta )=2$. Since $-1\leq \sin(\theta )\leq 1$, it is impossible
for $\sin(\theta )=2$. So, $\sin ^{-1}(2)$ is undefined.

Step 2:
For (b), we need to find the reference angle for ${\frac {-32\pi }{3}}$. If we add multiples of $2\pi$ to this angle, we get the
reference angle ${\frac {4\pi }{3}}$. So, $\sin \left({\frac {-32\pi }{3}}\right)=\sin \left({\frac {4\pi }{3}}\right)={\frac {-{\sqrt {3}}}{2}}$.

Step 3:
For (c), we need to find the reference angle for ${\frac {-17\pi }{6}}$. If we add multiples of $2\pi$ to this angle, we get the
reference angle ${\frac {7\pi }{6}}$. Since $\cos \left({\frac {7\pi }{6}}\right)={\frac {-{\sqrt {3}}}{2}}$, we have
$\sec \left({\frac {-17\pi }{6}}\right)=\sec \left({\frac {7\pi }{6}}\right)={\frac {2}{-{\sqrt {3}}}}={\frac {-2{\sqrt {3}}}{3}}$.

Final Answer:
a) undefined
b) ${\frac {-{\sqrt {3}}}{2}}$
c)${\frac {-2{\sqrt {3}}}{3}}$

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